OFFSET
1,1
FORMULA
a(n) = p+2*(number of solutions to x^y == y^x (mod p) where 1 < x < y < p). - Chai Wah Wu, Aug 30 2022
EXAMPLE
Solutions for a(3):
p x y x^y mod p y^x mod p
- - - --------- ---------
5 1 1 1 1
5 2 2 4 4
5 2 4 1 1
5 3 3 2 2
5 4 2 1 1
5 4 4 1 1
5 5 5 0 0
Total number of solutions: 7.
PROG
(Python)
from sympy import prime
def f(n):
S = 0
for x in range(1, n + 1):
for y in range(x + 1 , n + 1):
if ((pow(x, y, n) == pow(y, x, n))):
S += 2
return S + n
def a(n): return f(prime(n))
(Python)
from sympy import prime
def A355419(n): return (p:=prime(n))+sum(2 for x in range(2, p-1) for y in range(x+1, p) if pow(x, y, p)==pow(y, x, p)) # Chai Wah Wu, Aug 30 2022
(PARI) a(n) = my(p=prime(n)); sum(x=1, p, sum(y=1, p, Mod(x, p)^y == Mod(y, p)^x)); \\ Michel Marcus, Jul 05 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
DarĂo Clavijo, Jul 01 2022
STATUS
approved